Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX_V = 100010, MAX_E = MAX_V << 1;
int V, En;
int To[MAX_E], NextE[MAX_E], LastE[MAX_V];
const int MAX_ARR = MAX_V << 5;
int Tot;
int T[MAX_V], Cnt[MAX_ARR], Left[MAX_ARR], Right[MAX_ARR];
bool Vis[MAX_V];
int Size[MAX_V], Fa[MAX_V], Depth[MAX_V], MaxSon[MAX_V], Top[MAX_V];
int M;
vector<int> Peo[MAX_V];
inline void AddEdge(int u, int v) {
  En++;
  To[En] = v;
  NextE[En] = LastE[u];
  LastE[u] = En;
}
int Insert(int old, int val, int l = 1, int r = M) {
  int New = ++Tot;
  Cnt[New] = Cnt[old] + 1;
  if (l < r) {
    int m = ((l + r) >> 1);
    if (val <= m) {
      Left[New] = Insert(Left[old], val, l, m);
      Right[New] = Right[old];
    } else {
      Left[New] = Left[old];
      Right[New] = Insert(Right[old], val, m + 1, r);
    }
  }
  return New;
}
int Kth(int tu, int tv, int tlca, int tfa, int k) {
  int l = 1, r = M;
  while (l < r) {
    int lcnt =
        (Cnt[Left[tu]] + Cnt[Left[tv]] - Cnt[Left[tlca]] - Cnt[Left[tfa]]);
    if (lcnt >= k) {
      r = ((l + r) >> 1), tu = Left[tu], tv = Left[tv], tlca = Left[tlca],
      tfa = Left[tfa];
    } else {
      k -= lcnt, l = ((l + r) >> 1) + 1, tu = Right[tu], tv = Right[tv],
                 tlca = Right[tlca], tfa = Right[tfa];
    }
  }
  return l;
}
void DFS1(int u) {
  Vis[u] = 1, Size[u] = 1, Depth[u] = Depth[Fa[u]] + 1;
  T[u] = T[Fa[u]];
  for (vector<int>::iterator i = Peo[u].begin(); i != Peo[u].end(); ++i)
    T[u] = Insert(T[u], *i);
  for (int e = LastE[u]; e; e = NextE[e]) {
    int v = To[e];
    if (!Vis[v]) {
      Fa[v] = u;
      DFS1(v);
      Size[u] += Size[v];
      if (Size[MaxSon[u]] < Size[v]) MaxSon[u] = v;
    }
  }
}
void DFS2(int u) {
  Vis[u] = 1, Top[u] = u == MaxSon[Fa[u]] ? Top[Fa[u]] : u;
  for (int e = LastE[u]; e; e = NextE[e])
    if (!Vis[To[e]]) DFS2(To[e]);
}
int LCA(int u, int v) {
  while (Top[u] != Top[v])
    if (Depth[Top[u]] > Depth[Top[v]])
      u = Fa[Top[u]];
    else
      v = Fa[Top[v]];
  return Depth[u] < Depth[v] ? u : v;
}
inline void Main() {
  int Qn;
  scanf("%d%d%d", &V, &M, &Qn);
  for (int i = 1; i < V; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    AddEdge(u, v), AddEdge(v, u);
  }
  for (int i = 1; i <= M; i++) {
    int t;
    scanf("%d", &t);
    Peo[t].push_back(i);
  }
  DFS1(1);
  memset(Vis, 0, sizeof(Vis));
  DFS2(1);
  while (Qn--) {
    int v, u, a;
    scanf("%d%d%d", &v, &u, &a);
    int lca = LCA(v, u), tu = T[u], tv = T[v], tlca = T[lca], tfa = T[Fa[lca]];
    int K = (((Cnt[tu] + Cnt[tv] - Cnt[tlca] - Cnt[tfa])) < (a)
                 ? ((Cnt[tu] + Cnt[tv] - Cnt[tlca] - Cnt[tfa]))
                 : (a));
    printf("%d%c", K, " \n"[K <= 0]);
    for (int k = 1; k <= K; k++)
      printf("%d%c", Kth(tv, tu, tlca, tfa, k), " \n"[k == K]);
  }
}
int main() {
  Main();
  return 0;
}